squares

**Veronica1999** · January 16th 2012, 05:34 PM

I can't seem to figure out the logic behind this problem.
Would this problem have something to do with mode?
I know I am missing the main point of this problem.

**princeps** · January 16th 2012, 11:58 PM

Since $2^{n-12} +49$ is an odd number it must be perfect square also . So:

$x^2=2^{n-12}+49 \Rightarrow 2^{n-12}=x^2-49$

Now, if we make substitution $y=n-12$ we can write :

$2^y=x^2-49 \Rightarrow y= \log_{2} (x^2-49)$

This equality has integer solutions only if $x^2-49$ is power of $2$ so $x$ must be an odd number so we shall make substitution $x=2m-1$ and therefore :

$y=\log_{2} (4m^2-4m-48)=\log_{2}(4 \cdot(m^2-m-12))$

$=2+\log_{2} (m^2-4m+3m-12)=$

$=2+\log_{2} (m+3)(m-4)=2+\log_{2} (m+3)+\log_{2}(m-4)$

It is obvious that both $m+3$ and $m-4$ must be powers of $2$ and this is possible only if $m=5$ , therefore :

$y=2+3+0 \Rightarrow y=5 \Rightarrow n-12=5 \Rightarrow n=17 .$

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